1:

a: x/15=-2/6

=>x/15=-1/3

=>x=-5

b: 3/x=1,8/2

=>3/x=9/10

=>x=3*10/9=30/9=10/3

c: (x-3)/(x+2)=2/7

=>2x+4=7x-21

=>-5x=-25

=>x=5

d: (x+1)/3=(x-6)/8

=>8x+8=3x-18

=>5x=-26

=>x=-26/5

e: (2-x)/5=(x+4)/3

=>3(2-x)=5(x+4)

=>5x+20=6-3x

=>8x=-14

=>x=-7/4

g: (2x+1)/(-3)=(1-x)/2

=>2(2x+1)=3(x-1)

=>4x+2=3x-3

=>x=-5

a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

b, -4\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) < \(x\) < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))

   - \(\dfrac{13}{3}\).\(\dfrac{1}{3}\) < \(x\) < - \(\dfrac{2}{3}\).(-\(\dfrac{11}{12}\))

    - \(\dfrac{13}{9}\) < \(x\) < \(\dfrac{11}{18}\)

     \(x\) \(\in\) { -1; 0; 1}

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)          

a)4/5+x=2/3

x=2/3-4/5

x=-2/15

b)-5/6-x=2/3

x=-5/6-2/3

x=-3/2

c)1/2x+3/4=-3/10

1/2x=-3/10-3/4

1/2x=-21/20

x=-21/20:1/2

x=-21/10

d)x/3-1/2=1/5

x/3=1/5+1/2

x/3=7/10

10x/30=21/30

10x=21

x=21:10

x=21/10

\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)

\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)

\(x-2023=1\)

\(x=2024\)

Vậy..............

\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)

\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)

\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)=\dfrac{4}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)=\dfrac{4}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)=\dfrac{4}{21}\)

\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3}-\dfrac{1}{7}\right)=\dfrac{4}{21}\)

\(\Leftrightarrow x-2023=1\Leftrightarrow x=2024\)

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)

\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)

Em giải như XYZ olm em nhé

Sau đó em thêm vào lập luận sau:

\(x\) = \(\dfrac{11}{18}\)

Vì \(\in\) N* 

Vậy \(x\in\) \(\varnothing\)

Cảm ơn mn ạ.

\(TH1:x\ge0\)

\(\Rightarrow x\left(1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)

\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{1-\dfrac{5}{6}}=5\left(TM\right)\)

\(TH2:x< 0\)

\(\Rightarrow x\left(-1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)

\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{-1-\dfrac{5}{6}}=-\dfrac{5}{11}\left(TM\right)\)

Vậy ...

Giải:

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(TH1:x\ge0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

  \(x-\dfrac{5}{6}.x=\dfrac{5}{6}\)   

\(x.\left(1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

         \(x.\dfrac{1}{6}=\dfrac{5}{6}\) 

              \(x=\dfrac{5}{6}:\dfrac{1}{6}\) 

              \(x=5\) 

\(TH2:x\le0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(-x-\dfrac{5}{6}.x=\dfrac{5}{6}\) 

\(x.\left(-1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

     \(x.\dfrac{-11}{6}=\dfrac{5}{6}\) 

               \(x=\dfrac{5}{6}:\dfrac{-11}{6}\) 

               \(x=\dfrac{-5}{11}\) 

Vậy \(x\in\left\{\dfrac{-5}{11};5\right\}\)

\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)

\(\Leftrightarrow\dfrac{1680.\left(x-3\right)+1560.\left(x-3\right)-1456.\left(x-3\right)-1365.\left(x-3\right)}{21840}=0\)

\(\Leftrightarrow\left(x-3\right).\left(1680+1560-1456-1365\right)=0\)

\(\Leftrightarrow\left(x-3\right).419=0\)

\(\Leftrightarrow419x=1257\)

\(\Leftrightarrow x=3\)

Lời giải:

\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)

\((x-3)\left(\frac{1}{13}+\frac{1}{14}\right)=(x-3)\left(\frac{1}{15}+\frac{1}{16}\right)\)

\((x-3)\left[\left(\frac{1}{13}+\frac{1}{14}\right)-\left(\frac{1}{15}+\frac{1}{16}\right)\right]=0\)

Ta thấy:

\(\frac{1}{13}>\frac{1}{15}; \frac{1}{14}>\frac{1}{16}\Rightarrow \frac{1}{13}+\frac{1}{14}> \frac{1}{15}+\frac{1}{16}\)

Do đó biểu thức trong ngoặc vuông lớn hơn $0$ hay khác $0$

$\Rightarrow x-3=0$

$\Leftrightarrow x=3$

\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)

\(x-\dfrac{2}{3}\times x-6=1\)

\(x\times\left(1-\dfrac{2}{3}\right)=7\)

\(x\times\dfrac{1}{3}=7\)

\(x=21\)

\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)

\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)

\(15x-11=9+3x\)

\(12x=20\)

\(x=\dfrac{5}{3}\)

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)